Answer
Given: Right ‚ñ≥ABC where CD is an altitude of the triangle. We aim to prove that a^2 + b^2 = c^2. Since ‚ñ≥ABC and ‚ñ≥CBD share a right angle and angle B, the triangles are similar by AA. Similarly, ‚ñ≥ABC and ‚ñ≥ACD also have a right angle and angle A in common, confirming their similarity by AA. The proportions c/a and a/f hold true as they represent ratios of corresponding parts of similar triangles. These can be expressed as a^2 = cf and b^2 = ce. Adding b^2 to both sides of the first expression a^2 = cf yields a^2 + b^2 = cf + b^2. Substituting ce for b^2 allows us to write a^2 + b^2 = cf + ce. Utilizing the converse of the distributive property gives us a^2 + b^2 = c(f + e). The concluding statement of the proof is Because f + e = c, a^2 + b^2 = c^2.
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